Isomorphism of Measure Spaces

Definition

Let $X$ be a measurable space and $A$ a sigma algebra on $X$. Similary, Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y} be a measurable space and $B$ a sigma algebra on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y} . Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (X,A)} and $(Y,B)$ be measurable spaces.

• A map $f:X\rightarrow Y$ is called measurable if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}(B) \in A} for every $B\in B$.

• These two measurable spaces are called isomorphic if there exists a bijection $f:X\rightarrow Y$ such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and $f^{-1}$ are measurable (such $f$ is called an isomorphism).

Basic Theorem

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_1} and $X_{2}$ be Borel subsets of complete separable metric spaces. For the measurable spaces $(X_{1},B(X_{1}))$ and $(X_{2},B(X_{2}))$ to be isomoprhuic, it is necessary and sufficient that the sets $X_{1}$ and $X_{2}$ be of the same cardinality.

Properties

Smooth maps send sets of measure zero to sets of measure zero

Let $U$ be an open set of $\mathbb {R} ^{n}$, and let $f\colon U\rightarrow \mathbb {R} ^{n}$ be a smooth map. If $A\subset U$ is of measure zero, then $f(A)$ is of measure zero.

Mini-Sards Theorem

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U } be an open set of $\mathbb {R} ^{n}$, and let $f\colon U\rightarrow \mathbb {R} ^{m}$ be a smooth map. Then if $m>n$, $f(U)$ has measure zero in $\mathbb {R} ^{m}$.

Reference

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